Question: All sides of the convex pentagon $ABCDE$ are of equal length, and $\angle A = \angle B = 90^\circ$.  What is the degree measure of $\angle E$?
Answer: Because $AB=BC=EA$ and $\angle A = \angle B = 90^\circ$, quadrilateral $ABCE$ is a square, so $\angle AEC = 90^\circ$.

[asy]
pair A,B,C,D,G;
A=(0,10); B=(10,10);
C=(10,0); D=(5,-7.1);
G=(0,0);
draw(A--B--C--D--G--cycle,linewidth(0.8));
draw(G--C);
label("$A$",A,W);
label("$B$",B,E);
label("$C$",C,E);
label("$D$",D,S);
label("$E$",G,W);
[/asy]

Also $CD=DE=EC$, so $\triangle CDE$ is equilateral and $\angle CED =
60^\circ$.  Therefore \[
\angle E = \angle AEC + \angle CED =
90^\circ + 60^\circ = \boxed{150^\circ}.
\]